For each function , sketch the Bode asymptotic magnitude and asymptotic phase plots.

failure to yield the right-of-way to another vehicle or pedestrian is the primary Collision factor in about 20% of fatal and inj

laila [671]

You could just create a turn on green arrow. Have a button for pedestrians. The only way for them to turn is if they have a green arrow and the green arrow will only appear when pedestrians are stopped or finished walking and all cars are clear or fully stopped

8 0

3 years ago

Five batch jobs A through E arrive at a computer center in the order A to E at almost the same time. They have estimated running

Nikolay [14]

Answer:

Explanation:

The Turnaround time is the amount of time that elapses between the job arriving and completing. We assume that all jobs arrive at time 0, the turnaround time will simply be the time that they complete.

Round Robin:

we assume that the time quantum of the scheduler is 1 second.The table below gives a break down of which jobs will be processed during each time quantum. A asterisk(*) indicates that the job completes during that quantum.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

A B C D E A B C* D E A B D E A B D* E A B E A B* E A E A E* A A

C* = 8

D*=17

B*=23

E*=28

AVERAGE TURNAROUND = (8+17+23+28+30)/5 =106/5 = 21.2 MINUTES

B) PRIORITY SCHEDULING:

1-6 7-14 15-24 25-26 27-30

B E A C D

AVERAGETURNAROUND =(6+14+24+26+30)/5 = 100/5 = 20 MINUTES.

C)FCFS

1-10 11-16 17-18 19-22 23-30

A B C D E

AVERAGE TURNAROUND =(10+16+18+22+30)/5 = 96/5=19.2 MINTUES

D)SJF

1-2 3-6 7-12 13-20 21-30

C D B E A

AVERAGE TURNAROUND - (2+6+12+20+30)/5 =70/5 =14 MINUTES.

3 0

3 years ago

Individual/organizations who are responsible for monitoring, verifying, and sometimes certifying various aspects of the construc

Rus_ich [418]

The project or the site manager is responsible for monitoring or verifying the aspects of the construction process.

The site or the project manager is the individual that is responsible for the adherence to quality and guidance on the construction site.

The site or project manager has the following tasks

ensuring safety on the site
Monitoring the progress or the work that is being done
Sourcing materials
Solving problems

In conclusion, the project manager monitors and verifies aspects of the construction process.

8 0

2 years ago

A 50 Hz, four pole turbo-generator rated 100 MVA, 11 kV has an inertia constant of 8.0 MJ/MVA. (a) Find the stored energy in the

raketka [301]

Given Information:

Frequency = f = 60 Hz

Complex rated power = G = 100 MVA

Intertia constant = H = 8 MJ/MVA

Mechanical power = Pmech = 80 MW

Electrical power = Pelec = 50 MW

Number of poles = P = 4

No. of cycles = 10

Required Information:

(a) stored energy = ?

(b) rotor acceleration = ?

(c) change in torque angle = ?

(c) rotor speed = ?

Answer:

(a) stored energy = 800 Mj

(b) rotor acceleration = 337.46 elec deg/s²

(c) change in torque angle (in elec deg) = 6.75 elec deg

(c) change in torque angle (in rmp/s) = 28.12 rpm/s

(c) rotor speed = 1505.62 rpm

Explanation:

(a) Find the stored energy in the rotor at synchronous speed.

The stored energy is given by

$E = G \times H$

Where G represents complex rated power and H is the inertia constant of turbo-generator.

$E = 100 \times 8 \\\\E = 800 \: MJ$

(b) If the mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW, find rotor acceleration, neglecting mechanical and electrical losses.

The rotor acceleration is given by

$P_a = P_{mech} - P_{elec} = M \frac{d^2 \delta}{dt^2}$

Where M is given by

$M = \frac{E}{180 \times f}$

$M = \frac{800}{180 \times 50}$

$M = 0.0889 \: MJ \cdot s/ elec \: \: deg$

So, the rotor acceleration is

$P_a = 80 - 50 = 0.0889 \frac{d^2 \delta}{dt^2}$

$30 = 0.0889 \frac{d^2 \delta}{dt^2}$

$\frac{d^2 \delta}{dt^2} = \frac{30}{0.0889}$

$\frac{d^2 \delta}{dt^2} = 337.46 \:\: elec \: deg/s^2$

(c) If the acceleration calculated in part(b) is maintained for 10 cycles, find the change in torque angle and rotor speed in revolutions per minute at the end of this period.

The change in torque angle is given by

$\Delta \delta = \frac{1}{2} \cdot \frac{d^2 \delta}{dt^2}\cdot (t)^2$